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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: HHH(DDD)==0 is correct
Date: Sat, 5 Jul 2025 10:53:02 -0500
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On 7/5/2025 3:44 AM, Mikko wrote:
> On 2025-07-04 12:57:47 +0000, olcott said:
> 
>> On 7/4/2025 2:42 AM, Mikko wrote:
>>> On 2025-07-03 15:17:53 +0000, olcott said:
>>>
>>>> On 7/3/2025 9:50 AM, Richard Damon wrote:
>>>>> On 7/3/25 10:39 AM, olcott wrote:
>>>>>> On 7/3/2025 9:16 AM, Richard Damon wrote:
>>>>>>> On 7/2/25 10:50 PM, olcott wrote:
>>>>>>>> On 7/1/2025 11:37 AM, Mr Flibble wrote:
>>>>>>>>> On Mon, 30 Jun 2025 21:12:48 -0400, Richard Damon wrote:
>>>>>>>>>
>>>>>>>>>> On 6/30/25 2:30 PM, Mr Flibble wrote:
>>>>>>>>>>
>>>>>>>>>> PO just works off the lie that a correct simulation of the 
>>>>>>>>>> input is
>>>>>>>>>> different than the direct execution, even though he can't show 
>>>>>>>>>> the
>>>>>>>>>> instruction actually correctly simulated where they differ, 
>>>>>>>>>> and thus
>>>>>>>>>> proves he is lying.
>>>>>>>>>>
>>>>>>>>>> The closest he comes is claiming that the simulation of the 
>>>>>>>>>> "Call HHH"
>>>>>>>>>> must be different when simulated then when executed, as for "some
>>>>>>>>>> reason" it must be just because otherwise HHH can't do the 
>>>>>>>>>> simulation.
>>>>>>>>>>
>>>>>>>>>> Sorry, not being able to do something doesn't mean you get to 
>>>>>>>>>> redefine
>>>>>>>>>> it,
>>>>>>>>>>
>>>>>>>>>> You ar4e just showing you are as stupid as he is.
>>>>>>>>>
>>>>>>>>> No. A simulator does not have to run a simulation to completion 
>>>>>>>>> if it can
>>>>>>>>> determine that the input, A PROGRAM, never halts.
>>>>>>>>>
>>>>>>>>> /Flibble
>>>>>>>>
>>>>>>>> The most direct way to analyze this is that
>>>>>>>> HHH(DDD)==0 and HHH1(DDD)==1 are both correct
>>>>>>>> because DDD calls HHH(DDD) in recursive simulation and
>>>>>>>> DDD does not call HHH1(DDD) in recursive simulation.
>>>>>>>
>>>>>>> Nope. It seems you don't understand what the question actually IS 
>>>>>>> because you have just lied to yourself so much that you lost the 
>>>>>>> understanding of the queiston.
>>>>>>>
>>>>>>>>
>>>>>>>> *I can't imagine how Mike does not get this*
>>>>>>>
>>>>>>> I can't understand
>>>>>>>
>>>>>>>>
>>>>>>>> *Context of above dialogue*
>>>>>>>> *Context of above dialogue*
>>>>>>>> *Context of above dialogue*
>>>>>>>
>>>>>>> Context of your context:
>>>>>>>
>>>>>>> A Halt Decider is supposed to decide if the program given to it 
>>>>>>> (via some correct representation) will halt when run.
>>>>>>>
>>>>>>> Thus, "the input" needs to represent a program
>>>>>>>
>>>>>>>>
>>>>>>>> typedef void (*ptr)();
>>>>>>>> int HHH(ptr P);
>>>>>>>>
>>>>>>>> void DDD()
>>>>>>>> {
>>>>>>>>    HHH(DDD);
>>>>>>>>    return;
>>>>>>>> }
>>>>>>>>
>>>>>>>> int main()
>>>>>>>> {
>>>>>>>>    HHH(DDD);
>>>>>>>> }
>>>>>>>
>>>>>>> Which, by itself, isn't a valid input, or program. as HHH is 
>>>>>>> undefined.
>>>>>>>
>>>>>>> Each different definition of HHH, gives a different problem.
>>>>>>>
>>>>>>> Your "logic" seems to be based on trying to re-define what a 
>>>>>>> program is, which just makes it a lie.
>>>>>>>
>>>>>>> "Programs" must be complete and self-contained in the field of 
>>>>>>> computability theory, something you don't seem to understand.
>>>>>>>
>>>>>>>>
>>>>>>>> Termination Analyzer HHH simulates its input until
>>>>>>>> it detects a non-terminating behavior pattern. When
>>>>>>>> HHH detects such a pattern it aborts its simulation
>>>>>>>> and returns 0. (HHH1 has identical code)
>>>>>>>>
>>>>>>>>
>>>>>>>
>>>>>>> But it CAN'T simulate the above input. as it isn't valid.
>>>>>>>
>>>>>>> You need to add the code of HHH to the input to let HHH simulate 
>>>>>>> "the input" to get anything.
>>>>>>>
>>>>>>
>>>>>> No I do not. The above paragraph has every detail that is needed.
>>>>>
>>>>> Then how do you correctly simulate something you do not have.
>>>>>
>>>>> Note, your "description" of HHH is just incorrect, as it is also 
>>>>> incomplete.
>>>>>
>>>>> Simulating a LIE just gives you a lie.
>>>>>
>>>>>>
>>>>>>> And at that point, you have different inputs for different HHHs, 
>>>>>>> and possibly different behaviors, which you logic forgets to take 
>>>>>>> into account, which just breaks it.
>>>>>>>
>>>>>>
>>>>>> Wrong.
>>>>>> It is because the what I specified does take this
>>>>>> into account that HHH(DDD)==0 and HHH1(DDD)==1 are correct.
>>>>>
>>>>> Nope, becausee it violates the DEFINITION of what it means to 
>>>>> simulate something.
>>>>
>>>> *You don't even know what you mean by this*
>>>> What I mean is the execution trace that is derived
>>>> within the semantics of the C programming language.
>>>
>>> C lanbuage definition does not specifiy the senatics of the non-standard
>>> lanugage extension that your HHH and HHH1 use.
>>
>> *This is the ONLY specification of HHH that chatbots see*
> 
> That is irrelevant. What is relevant is the specification or lack of
> that readers of comp.theory see.
> 

Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.

That is what they and the Chatbots both saw proving
that claims it was not sufficient were gaslighting.

All the Chatbots figured out on their own that DDD
correctly simulated by HHH specifies recursive simulation
such that DDD cannot possibly reach its own simulated
"return" statement. They only had the above paragraph
(and the code) as their basis.

*ChatGPT, Gemini, Grok and Claude all agree*
DDD correctly simulated by HHH cannot possibly reach
its simulated "return" statement final halt state.

https://chatgpt.com/share/685ed9e3-260c-8011-91d0-4dee3ee08f46
https://gemini.google.com/app/f2527954a959bce4
https://grok.com/share/c2hhcmQtMg%3D%3D_b750d0f1-9996-4394-b0e4-f76f6c77df3d 
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